# Where characteristic enters a resolution

From a six-vertex projective plane to two different Betti numbers.

2026-10-17 · Study note · chain calculation

> An assistant-prepared reconstruction of a classical example, with independent Python/SymPy checks. The general Hochster formula is invoked, not claimed as independently proved in the saved learning record.

## The question in my saved reading

In the monomial-ideal reading I asked how a generating set leads to an exact chain and how one knows that a list generates the whole kernel. The projective-plane example here is an assistant-prepared continuation of that question, not a claim that I had already proved Hochster’s formula.

## What changes in the comparison

The integer boundary matrix stays fixed; changing coefficients can change its rank. The ideal and its quotient must be distinguished when translating homology into Betti indices. The general formula’s proof remains separate from using it in this example.

## Fix the complex before the field

Let Δ have vertices 1,…,6 and the ten facets listed below. Every edge occurs in two triangles. This is the standard six-vertex triangulation of the real projective plane; its one-skeleton is the complete graph K₆. The complete graph alone is not the full triangulation: the choice of triangles matters.

$$
123,124,135,146,156,236,245,256,345,346.
$$

Orient every simplex by increasing vertex order. The chain groups over a field k have dimensions 6, 15, and 10. The signed boundary removes one vertex at a time with alternating signs.

$$
\partial_2[i,j,\ell]=[j,\ell]-[i,\ell]+[i,j],\qquad\partial_1[i,j]=[j]-[i].
$$

$$
k^{10}\xrightarrow{\partial_2}k^{15}\xrightarrow{\partial_1}k^6,\qquad\partial_1\partial_2=0.
$$

## The same integer matrix, different ranks

Compute over Q using exact integer elimination, or reduce the signed matrix modulo a prime and use finite-field elimination. There is no floating-point tolerance. Connectivity gives rank ∂₁ = 5 in either case.

| Field | rank ∂₂ | dim H₁ | dim H₂ |
| --- | --- | --- | --- |
| Q | 10 | 0 | 0 |
| F₂ | 9 | 1 | 1 |

$$
\begin{aligned}\dim H_1(\Delta;k)&=15-\operatorname{rank}\partial_1-\operatorname{rank}\partial_2,\\\dim H_2(\Delta;k)&=10-\operatorname{rank}\partial_2.\end{aligned}
$$

See the website for the interactive calculation; exact seed scripts and results are provided alongside it.

The nonzero Smith diagonal of the integer second boundary is nine 1s followed by a 2. Together with the first boundary, this gives H₁(Δ;Z) ≅ Z/2 and H₂(Δ;Z) = 0. The universal coefficient theorem explains both mod-two homology groups: H₁ receives the tensor contribution and H₂ receives the Tor contribution from integral H₁.

## From topology to a graded Betti number

Put S = k[x₁,…,x₆] and let IΔ be the Stanley–Reisner ideal generated by squarefree monomials on nonfaces. Hochster’s formula for the quotient k[Δ] = S/IΔ is:

$$
\beta_{i,j}^{S}(k[\Delta])=\sum_{\substack{W\subseteq[6]\\|W|=j}}\dim_k\widetilde H_{j-i-1}(\Delta_W;k).
$$

At j = 6 there is only one vertex subset. H₂ contributes to i = 3, and H₁ contributes to i = 4. Consequently the full-vertex-degree entries increase by one in characteristic two:

$$
\begin{aligned}\beta_{3,6}^{\mathbb F_2}(S/I_\Delta)-\beta_{3,6}^{\mathbb Q}(S/I_\Delta)&=1,\\\beta_{4,6}^{\mathbb F_2}(S/I_\Delta)-\beta_{4,6}^{\mathbb Q}(S/I_\Delta)&=1.\end{aligned}
$$

These indices refer to S/IΔ. For the ideal itself, βᵢⱼ(IΔ) = βᵢ₊₁,ⱼ(S/IΔ) for i ≥ 0. Confusing the ideal with its quotient shifts the homological index.

## Why powers are a real next problem

IΔ² is usually not squarefree. The displayed Hochster formula for Δ cannot simply be reused with an exponent attached. Polarization can replace a monomial ideal by a squarefree ideal in more variables while preserving graded Betti numbers; its complex must actually be constructed.

A concrete project should choose a controlled subdivision family, compare its first few powers, and then seek a mechanism that explains propagation or disappearance of the characteristic dependence. A finite table is evidence for that mechanism, not a proof for every power.

## Sources

- [Miller & Perkinson — Eight Lectures on Monomial Ideals](https://sites.math.duke.edu/~ezra/Queens/cocoa.pdf)

- [Bolognini et al. — Powers of monomial ideals with characteristic-dependent Betti numbers](https://link.springer.com/article/10.1007/s40687-022-00318-2)